题目链接
暴力 时间复杂度为O(n^2)
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| #include<iostream>
using namespace std;
const int N = 100010;
int q[N];
int n;
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
scanf("%d", &q[i]);
int cnt = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < i; j++)
{
if (q[j] > q[i])
cnt++;
}
}
cout << cnt;
return 0;
}
|
归并
时间复杂度为O(nlogn)，又印证了陈越姥姥的那句话了。
归并的思想也是分而治之。递归不能忘了写递归出口啊。
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| #include<iostream>
using namespace std;
const int N = 100010;
int q[N], tmp[N];
int n;
long long res;
void merge_sort(int q[], int l, int r)
{
if (l >= r)
return;
int mid = l + r >> 1;
merge_sort(q, l, mid);
merge_sort(q, mid + 1, r);
int i = l, j = mid + 1;
int k = 0;
while (i <= mid && j <= r)
{
if (q[i] <= q[j])
tmp[k++] = q[i++];
else
{
res += mid - i + 1;
tmp[k++] = q[j++];
}
}
while (i <= mid)
tmp[k++] = q[i++];
while (j <= r)
tmp[k++] = q[j++];
for (int i = l, j = 0; i <= r; i++, j++)
q[i] = tmp[j];
}
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
scanf("%d", &q[i]);
merge_sort(q, 0, n - 1);
cout << res;
system("pause");
return 0;
}
|
